# Lagrange Multipliers and Inequalities

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• Jul 22nd 2010, 03:19 PM
fobos3
Lagrange Multipliers and Inequalities
Is there a way to apply the method of Lagrangian Multipliers when your equation of constraint is an inequality(it won't be called an equation of constraint then I guess). Here is an example of what I came up with.

Say $\displaystyle f(x)=x^2$

You want to find the extrema satisfying the inequality $\displaystyle -1\le x\le1$

If you introduce a variable $\displaystyle z$ that satisfies the equation $\displaystyle x^2+z^2=1$ that means $\displaystyle x\in[-1,1]$

$\displaystyle g(x,z)=x^2+z^2=1$ becomes your equation of constraint. Now we have

$\displaystyle \nable f=\lambda\nabla g$ where $\displaystyle \nabla =\left(\dfrac{\partial}{\partial x},\dfrac{\partial}{\partial z}\right)$

We get $\displaystyle \left(2x,0\right)=\lambda\left(2x,2z\right)$

From $\displaystyle 2\lambda z=0$ we get $\displaystyle z=0$ or $\displaystyle \lambda=0$

From $\displaystyle 2x=2\lambda x$ we get $\displaystyle x=0$ or $\displaystyle \lambda=1$

For $\displaystyle x=0$ we get our minimum.

For $\displaystyle \lambda=1$ we have $\displaystyle z=0$ and $\displaystyle x^2=1\therefore x=\pm 1$ which is the local maximum.

Now what I want to know is, if there is a way to solve for more general inequalities like $\displaystyle x<3$ or $\displaystyle x>7$.

Any input will be appreciated.
• Jul 24th 2010, 06:38 AM
Media_Man
The First Dimension
In principle, I think the answer is yes, by introducing the simple disk as you have, which was a good idea. So for $\displaystyle x<3$, you would let $\displaystyle g(x,z)=x^2+z^2=9$ . For $\displaystyle x>7$, the domain is a bit trickier, but you can replace with $\displaystyle \frac1x<\frac17$ and make $\displaystyle g(x,z)=\frac1{x^2}+z^2=\frac1{49}$ to get $\displaystyle x>7$ or $\displaystyle x<-7$. I think this might be what you are looking for.

However, keep in mind that in one dimension, there is little point to Lagrange multipliers. Consider: if $\displaystyle \nabla f = \lambda \nabla g$, then by letting $\displaystyle \lambda=0$, $\displaystyle \nabla f=0$, so $\displaystyle (\frac{\partial}{\partial x}, 0)=(0,0)$, i.e. $\displaystyle f'(x)=0$, which is the standard one-dimensional way of finding the maxima/minima anyway, having nothing to do with $\displaystyle g(x,z)$.

You might try experimenting with your method on functions $\displaystyle f(x,y)$ over two variables to see if there is something truly new to be discovered.